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Find out the energy of H atom in first excitation state. The value of permitivity factor $4\pi E_{\large\circ} = 1.11264\times10^{-10} C^2N^{-1}m{-2}$

$(a)\;2.443\times10^{-19}J\qquad(b)\;5.443\times10^{-19}J\qquad(c)\;5.443\times10^{-10}J\qquad(d)\;3.432\times10^{-19}J$

1 Answer

$E_n = \large\frac{2\pi^2Z^2me^4}{(4\pi E_{\large\circ})^2n^2h^2}$
$= \large\frac{2\times(3.14)^2\times1^2\times9.108\times10^{-30}(1.602\times10^{-19})^4}{(1.11264\times10^{-10})^2\times2^2\times(6.625\times10^{-34})^2}$
$=5.44\times10^{-19} $Joule
answered Jan 31, 2014 by sharmaaparna1
 

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